Question

In $YDSE$, the thickness of a glass slab $(\mu =1.5)$ which should be placed in front of the upper slit $S_1$ so that the central maximum now lies at a point where the $5^{th}$ bright fringe was lying earlier (before inserting the slab), is given by $x\times {10}^4\dot{\text{A}}$. The value of $x$ is (Integer only)
[Wavelength of light $=5000\dot{\text{A}}$]

Answer 1

The formula for fringe shift is given by,

$\Delta y=\frac{\beta }{\lambda }(\mu -1)t$

According to the question,
$\Delta y=5\beta$

$\Rightarrow \frac{(\mu -1)t\beta }{\lambda }=5\beta$

$\Rightarrow t=\frac{5\lambda }{\mu -1}=\frac{25000}{1.5-1}$

$\Rightarrow t=50000\dot{\text{A}}=5\times {10}^4\dot{\text{A}}$

Given, $t=x\times {10}^4\dot{\text{A}}$

$\therefore x=5$